Find Resistance Between A and B - 5.0Ω, 98.0Ω, 16.0Ω

[n77ler]

Homework Statement

A number of resistors of values R1 = 5.0 Ω, R2 = 98.0 Ω, and R3 = 16.0 Ω are connected as shown in the figure. What is the resistance between points A and B?

Homework Equations

The Attempt at a Solution

Ok so the 4 R_1's and R_3 are in series on the end which gives a resistance of 36.0 ohm. This resistance is in parallel with R_2 so it's 1/ (1/first resistance + 1/R2 ) =26.3ohm. The 4 R1's and R2 are in series so its R1(4)+R2 =118. Now this resistance is in parallel with the first R2 so its 1/ (1/118 +1/98)= 53.5ohm + the 2 R1's at the start = 63.5

So 63.5 ohm + 26.3 ohm = 89.8 ohm but this isn't right. Can someone help me out please.

 

[Doc Al]

Ok so the 4 R_1's and R_3 are in series on the end which gives a resistance of 36.0 ohm. This resistance is in parallel with R_2 so it's 1/ (1/first resistance + 1/R2 ) =26.3ohm. The 4 R1's and R2 are in series so its R1(4)+R2 =118.

Redo that last step. The 4 R1s are in series with the effective resistance calculated in the previous step, not with R2 alone.

 

[n77ler]

so R1(4) +R2 + 26.3 ohm= 144.3 ohm, so then do I just follow the steps I have already used?

 

[Doc Al]

so R1(4) +R2 + 26.3 ohm= 144.3 ohm, so then do I just follow the steps I have already used?

No, you're still making an error. As you work from right to left, you replace resistors with their equivalent resistance.

(1) You replaced 4R1 + R3 with their equivalent = 36 ohms
(2) Now R2 & 36 ohms are in parallel, replace them with their equivalent = 26.3 ohms
(3) Now you have 4R1 in series with 26.3 ohms (not R2; R2 is gone!)
... and so on

 

[n77ler]

Oh ok ,I understand it good now, thank- you very much!
26.3ohm+4R1= 46.3
Parallel with R2 so 1/ (1/98+1/46.3)= 31.44ohm + 2R1 on the very left = 41.4 ohm :)

 

[Doc Al]

You got it.

 

[n77ler]

Ok so there's another part of the question I'm having trouble with I worked through it but its the first time I've seen anything like it.

A-C was I1, C-D was I3 D-B was I1

I1-I3=0
I3-I1=0
I1-I3=I3-I1
2 I1-2 I3=0
I3=I1
E-I3R-I1R=0
E=I1R-I1R=0
E-2I1R=0
E= (2I1R)/(2R) I1=E/2R= 110.0V/2(98.0ohm) = 0.561A

 

[Doc Al]

Ok so there's another part of the question I'm having trouble with I worked through it but its the first time I've seen anything like it.

A-C was I1, C-D was I3 D-B was I1

I1-I3=0

At junction C (or any other junction) it's the total current flow that must add to zero. You counted the current from the A-C (I1) and D-C (I3), but what about the branch going to the right out of C?

What are you given for this part of the problem? The total voltage across A-B? If so, do the opposite of what you did before. This time work backwards from left to right, using the results of your previous work to find the various currents.

 

[n77ler]

So are my working relevant at all? Or do I need to scrap them completely and start over?

 

[Doc Al]

Start over. Your first equation, I1-I3=0, is incorrect.

 

[n77ler]

kk I've got work so I can't solve now, Ill be back later, you may not be here but thankyou very much for the help today!

 

[n77ler]

So will my first equation have to consist of I1,I2,I3,I4,I5,I6 because of the complexity of the circuit?